Rays Mx and Ny are parallel and point P lies between the two lines, with angle xMP = 110 degrees and angle yNP = 140 degrees. Find angle MPN.

Step-by-step solution

1. 1

   Draw an auxiliary line through P parallel to both. The two given angles sit at different vertices, so nothing links them yet. Introducing a ray $Pz$ through $P$ parallel to $Mx$ (and hence to $Ny$) splits $\angle MPN$ into two pieces, each of which can be tied to one of the given angles.

2. 2

   Use co-interior angles at M. $Mx\parallel Pz$ with $MP$ as the transversal, so $\angle xMP$ and $\angle MPz$ are co-interior and sum to $180^{\circ}$:

   $\angle MPz=180^{\circ}-110^{\circ}=70^{\circ}$

3. 3

   Use co-interior angles at N. Likewise $Ny\parallel Pz$ with $NP$ as the transversal:

   $\angle NPz=180^{\circ}-140^{\circ}=40^{\circ}$

4. 4

   Add the two pieces. Because $Pz$ lies inside $\angle MPN$ (that is what "$P$ lies between the two lines" guarantees),

   $\angle MPN=\angle MPz+\angle NPz=70^{\circ}+40^{\circ}=110^{\circ}$

   $\boxed{110^{\circ}}$

5. 5

   Note the shortcut and its validity. The same computation is the one-liner

   $\angle MPN=360^{\circ}-110^{\circ}-140^{\circ}=110^{\circ}$

   because the three angles $\angle xMP$, $\angle MPN$ and $\angle yNP$ close up around the auxiliary line. This shortcut applies whenever the zig-zag point genuinely lies between the two parallels; if it lay outside the strip, the two pieces would be subtracted instead of added.

6. 6

   Sanity-check the size. Both given angles are obtuse, so each contributes a modest acute piece at $P$ ($70^{\circ}$ and $40^{\circ}$), and their sum $110^{\circ}$ is obtuse but less than $180^{\circ}$ — consistent with $P$ sitting strictly between the lines.